document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=777335>Question 777335</A>: <I><SPAN STYLE=\"word-break: break-all;\"> to make gunmetal, copper, tin and zinc are used in the ratio 43:5:2. what quantities of tin and zinc are used with 86kg of copper.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #474069 by </B><A HREF=/tutors/aboutme.mpl?userid=Cromlix><B>Cromlix(4381)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Cromlix><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Cromlix><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=474069\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Ratio 43:5:2<BR>\n" );
document.write( "Copper:Zinc:Tin<BR>\n" );
document.write( "If 86kg of copper used<BR>\n" );
document.write( "10kg of Zinc and 4kg of Tin<BR>\n" );
document.write( "would be required to stay <BR>\n" );
document.write( "within the ratio required.<BR>\n" );
document.write( "Hope this helps.<BR>\n" );
document.write( ":-) </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );